hdu多校第3场C. Dynamic Graph Matching

Problem C. Dynamic Graph Matching
 
Time Limit: / MS (Java/Others)    Memory Limit: / K (Java/Others)
Total Submission(s):     Accepted Submission(s): 
 
Problem Description
In the mathematical discipline of graph theory, a matching in a graph is a set of edges without common vertices.
You are given an undirected graph with n vertices, labeled by ,,...,n. Initially the graph has no edges.
There are  kinds of operations :
+ u v, add an edge (u,v) into the graph, multiple edges between same pair of vertices are allowed.
- u v, remove an edge (u,v), it is guaranteed that there are at least one such edge in the graph.
Your task is to compute the number of matchings with exactly k edges after each operation for k=,,,...,n2. Note that multiple edges between same pair of vertices are considered different.
 
Input
The first line of the input contains an integer T(≤T≤), denoting the number of test cases.
In each test case, there are  integers n,m(≤n≤,nmod2=,≤m≤), denoting the number of vertices and operations.
For the next m lines, each line describes an operation, and it is guaranteed that ≤u<v≤n.
 
Output
For each operation, print a single line containing n2 integers, denoting the answer for k=,,,...,n2. Since the answer may be very large, please print the answer modulo +.
 
Sample Input
 
+
+
+
+
-
-
+
+  
 
Sample Output
 
Source
 Multi-University Training Contest 
 
Recommend
chendu
 
Statistic | Submit | Discuss | Note

装压dp

+：操作很简单就是：dp[i]+=dp[i-(1<<u)-(1<<v)];

-：操作就想象成反的： dp[i]-=dp[i-(1<<u)-(1<<v)];拿总的减去用到用到u,v

类似于背包某个物品不能放；

#include <cstdio>
#include <cstring>
#include <iostream>
//#include <algorithm>
#include <vector>
using namespace std;
#define ll long long
//#define mod 998244353
const int mod=1e9+;
 
ll dp[<<];//dp[i]:i集合内点完全匹配的方案数
int bit[<<];//i的二进制的1个数；
ll ans[];
int lowbit(int x) {return x&-x;}
int calc(int x)
{
    int res=;
    while(x){res++;x=x-lowbit(x);}
    return res;
}
int main()
{
    for(int i=;i<(<<);i++)
        bit[i]=calc(i);
    int T,n,m,u,v;
    char op[];
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        memset(dp,,sizeof dp);
        memset(ans,,sizeof ans);
        dp[]=;
        while(m--)
        {
            scanf("%s%d%d",op,&u,&v);
            u--;v--;
            if(op[]=='+')
            {
                for(int i=(<<n)-;i>;i--)
                if(((<<u)&i)&&((<<v)&i))
                {
                    dp[i]+=dp[i-(<<u)-(<<v)];
                    ans[bit[i]]+=dp[i-(<<u)-(<<v)]; //对于i集合所有点的匹配，会对ans造成的影响
                    dp[i]=dp[i]%mod;
                    ans[bit[i]]=ans[bit[i]]%mod;
                }
            }
 
            else
            {
                for(int i=;i<<<n;i++)
                if(((<<u)&i)&&((<<v)&i))
                {
                    dp[i]-=dp[i-(<<u)-(<<v)];
                    ans[bit[i]]-=dp[i-(<<u)-(<<v)];
                    dp[i]=(dp[i]+mod)%mod;
                    ans[bit[i]]=(ans[bit[i]]+mod)%mod;
                }
            }
            for(int i=;i<=n;i=i+)
            {
                if(i!=)
                    cout<<" ";
                printf("%lld",ans[i]);
 
            }
            cout<<endl;
 
        }
 
    }
 
    return ;
}